A reaction is represented by this equation: \(2 \mathrm{W}(a q) \rightleftharpoons \mathrm{X}(a q)+2 \mathrm{Y}(a q) \quad K_{c}=5 \times 10^{-4}\) (a) Write the mathematical expression for the equilibrium constant. (b) Using concentrations of \(\leq 1 M,\) identify two sets of concentrations that describe a mixture of \(\mathrm{W}, \mathrm{X},\) and \(\mathrm{Y}\) at equilibrium.

Short Answer

Expert verified
The expression for the equilibrium constant is \( K_c = \frac{[X][Y]^2}{[W]^2} \). Two sets of equilibrium concentrations using \( K_c = 5 \times 10^{-4} \) could be [W] = 1 M, [X] = 5 M, [Y] = 0.01 M or [W] = 0.1 M, [X] = 0.5 M, [Y] = 0.1 M.

Step by step solution

01

- Write the Expression for the Equilibrium Constant

For the equilibrium reaction, we write the expression for the equilibrium constant, Kc, by using the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients. For the given reaction, the equilibrium expression is: \( K_c = \frac{[X][Y]^2}{[W]^2} \)
02

- Determine Sets of Concentrations at Equilibrium

To describe mixtures of W, X, and Y at equilibrium using Kc, we can choose values for the concentrations of W, X, and Y such that when substituted in the equilibrium expression, the result is equal to 5 x 10^-4. For the first set, we can assume a concentration of W to be 1 M and then calculate the concentrations of X and Y. For the second set, we'll assume a different concentration of W and repeat the calculation.
03

- Calculate the First Set of Concentrations

Let's assume [W] is 1 M at equilibrium, and then solve for [X] and [Y]. When [W] = 1 M, we have \( K_c = \frac{[X][Y]^2}{(1)^2} = 5 \times 10^{-4} \). Let's choose [Y] = 0.01 M, then \( [X] = K_c \div [Y]^2 = 5 \times 10^{-4} \div (0.01)^2 = 5 \). So the first set of concentrations at equilibrium would be [W] = 1 M, [X] = 5 M, and [Y] = 0.01 M.
04

- Calculate the Second Set of Concentrations

For another set of concentrations, let's choose [W] = 0.1 M. Then, we solve for [X] and [Y] such that \( K_c = \frac{[X][Y]^2}{(0.1)^2} \). If we choose [Y] = 0.1 M, we find \( [X] = K_c \div [Y]^2 = 5 \times 10^{-4} \div (0.1)^2 = 0.5 \). This gives us the second set of equilibrium concentrations: [W] = 0.1 M, [X] = 0.5 M, and [Y] = 0.1 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant Expression
Understanding the equilibrium constant expression is critical when studying chemical equilibrium. It is a way to quantify the extent of a reaction, or in simple terms, to tell us how far a reaction will go before reaching equilibrium. The equilibrium constant, represented by Kc for reactions in solution or Kp for reactions involving gases, is calculated from the concentrations (or partial pressures) of the products and reactants.

For a reaction at equilibrium, the expression is derived by taking the product of the concentrations of the products, each raised to the power of their stoichiometric coefficient, and dividing this by the product of the concentrations of the reactants, each also raised to their stoichiometric coefficient. This is symbolized in the equation \[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \] In practice, this means that if a reaction has reached equilibrium at a certain temperature, the ratio of the concentration of the products to reactants will remain constant, represented by the equilibrium constant.
Stoichiometry
Stoichiometry is the section of chemistry that involves using relationships between reactants and products in a chemical reaction to predict quantities of either reactants or products. It's like a mathematical recipe for a chemical reaction telling us how much of each substance we need to make a certain amount of another substance or how much we produce from a given amount of reactants.

The stoichiometric coefficients are the numbers in front of the chemical species in a balanced chemical equation and indicate the ratios in which reactants combine and products form. When applying stoichiometry to equilibria, these coefficients are crucial as they are used as exponents in the equilibrium expression. They tell you how the concentration of each reactant and product influences the position of equilibrium.
Equilibrium Concentration Calculation
Calculating the equilibrium concentration of substances in a reaction involves using the equilibrium constant expression along with the stoichiometry of the reaction. By understanding the chemical equation and the value of the equilibrium constant, we can calculate unknown concentrations at equilibrium given some concentrations.

It's a bit like a puzzle where you must use clues (the given concentrations and Kc) to find the missing pieces (the unknown concentrations). This involves algebraic manipulations, where you substitute known values into the equilibrium expression and solve for the unknowns. Having the right stoichiometry for each substance is key here, as any mistake will lead to incorrect results.
Reversible Reactions
Reversible reactions are those that can proceed in both forward and reverse directions. In other words, the products can re-form the original reactants. When a reversible reaction is allowed to proceed in a closed system without interference, it will eventually reach a state of balance, or equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction.

This does not mean that the reaction has stopped; rather, the concentrations of reactants and products remain constant over time. It’s a dynamic balance, as reactants continue to form products at the same rate that products break down into reactants. This concept greatly influences how we predict the behavior of a reaction and calculate the equilibrium concentrations of the involved substances.

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Most popular questions from this chapter

When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?

If a reaction is reversible, when can it be said to have reached equilibrium?

Acetic acid is a weak acid that reacts with water according to this equation: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(a q) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)\) Will any of the following increase the percent of acetic acid that reacts and produces \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) ion? (a) Addition of HCl (b) Addition of NaOH (c) Addition of \(\mathrm{NaCH}_{3} \mathrm{CO}_{2}\)

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